The difference between the least common multiple and the greatest common multiple of a and B is 20, which is equal to the sum of the two numbers

The difference between the least common multiple and the greatest common multiple of a and B is 20, which is equal to the sum of the two numbers

Let the greatest common divisor be m, then let a = am, B = BM, where a and B are coprime
So am + BM = 20, a + B = 20 / m
Abm-m = 20, AB-1 = 20 / m
So AB-1 = a + B, so AB-A = B + 1, so a = (B + 1) / (B-1) = 1 + 2 / (B-1), so it can only be a = 2, B = 3, or a = 3, B = 2
So the two numbers are 20 / (2 + 3) * 2 = 8 and 20 / (2 + 3) * 3 = 12

A two digit number, which can be divided by 2, has a divisor 5, and is a multiple of 3. What is the maximum number of this two digit number

90 2*3*3*5

A number contains both the divisor 24 and the multiple of 36. What is the minimum number

Least common multiple of 24 and 36: 72
The minimum number is 72

The maximum divisor of a number is ten 36, which is the minimum multiple of all its divisors

This number is the smallest multiple of (36) of all his divisors (1, 2, 3, 4, 6, 9, 12, 18, 36)

We know the equation x & sup2; - (K + 2) x + 2K = 0 (1) about X. we prove that no matter K takes any real number, the equation always has real root
(2) If one side of the isosceles triangle ABC a = 3, and the lengths B and C of both sides are exactly the two roots of the equation, find the perimeter of △ ABC
Quick - give me extra points

1. Because X & # 178; - (K + 2) x + 2K = 0, then (X-2) * (x-k) = 0 leads to x = 2 or x = k, so no matter K takes any real number, the equation always has real roots. 2. From the above, we can see that the two roots of the equation are 2 or K. because it is an isosceles triangle, the two cases (1) if 2 is the bottom, then k = a = 3, perimeter is 3 + 3 + 2 = 8 (2) a = 3

What is the value range of | Z | if the complex Z satisfies | Z-I | = 2

In the rectangular coordinate system, Z is a circle with I as the center and 2 as the length

α belongs to [0, Pai / 2], find the value range of modulus of complex z = 1-sin α + I * 2cos α

The square of Z | Z | ^ 2 = (1-sin α) ^ 2 + 4cos α ^ 2, because α belongs to [0, PI / 2], in this interval, the upper chord function is monotone, so the value range of sin α is [0,1]. Let x = sin α, then the original formula is reduced to: | Z | ^ 2 = - 3x ^ 2-2x + 5, and its axis of symmetry is x = - 1 / 3, so x belongs to [0,1] and should be on the right side of the axis of symmetry

Let z = 2 + cosa + isina, a belong to [0180], w = I + 1, find the value range of | z-w |
Such as the title

Z is the upper part of a circle whose center is [2,0] and radius is 1
W is the point (1,1)
It can be obtained by drawing
The shortest is (radical 2) - 1 and the longest is radical 5

Given the complex z = 1 + I, the module of complex (z2-3z + 6) / (Z + 1) is

(z²-3z+6)/(z+1)
＝[(1+i)²-3(1+i)+6]/(1+i+1)
＝(3-i)/(2+i)
＝1-i,
∴|(z²-3i+6)/(z+1)|
＝|1-i|
＝√[1²+(-1)²]
＝√2.

Let z = 1 + 2I, then z2-5 / z =?

If z = 1 + 2I, then
z²=1+4i-4=4i-3
5/z＝5/（1＋2i）=5(1-2i)/(1+2i)（1-2i)=5(1-2i)/5=1-2i
z²－5/z＝4i-3-（1-2i）=-4+6i